Find $\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}$ I am just trying to calculate 
$$\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}.$$
To do this I apply formula for sum of fourth powers of $n$ number. My result: $$\lim_{n\to\infty}\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}=1$$. I'm intrested in finding other method to solve the following problem.  
 A: Let $\ell\in \mathbb R$ be a finite number. If
$$\lim _{n\to \infty }{\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}=\ell ,\ $$
then, the limit
$${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\ell .\ }$$
This is the Stolz–Cesàro theorem. In your case
$$a_n=\sum_{k=1}^n k^4, \quad b_n=\sum_{k=1}^{n+1} k^4$$
and
$$a_{n+1}-a_n=(n+1)^4, \quad b_{n+1}-b_n=(n+2)^4,$$
so the limit is equal to 1.
A: As remarked by SBA, by denoting $1^4+2^4+\ldots+n^4$ as $p(n)$ we have
$$ \lim_{n\to +\infty}\frac{p(n)}{p(n+1)} \stackrel{\text{Stoltz-Cesàro}}{=}\lim_{n\to +\infty}\frac{n^4}{(n+1)^4} = 1 $$
but the same conclusion simply follows from the observation that $p(n)$ is a polynomial.
A: Hint:
$$\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}
=\frac{1^4+2^4+\dots+n^4+(n+1)^4-(n+1)^4}{1^4+2^4+\dots+n^4+(n+1)^4}
$$
A: If
$a_n > 0,
s_n = \sum_{k=1}^n a_n,
r_n = s_{n+1}/s_n$,
then,
if $a_{n+1}/a_n \to 1$,
$r_n \to 1$.
Proof:
$r_n-1
=a_{n+1}/s_n
$.
Since 
$a_{n+1}/a_n \to 1$,
then,
for any fixed $m$,
$a_{n+k}/a_n \to 1$
for $1 \le k \le m$.
Therefore,
for any fixed $m$,
$\dfrac{a_{n+1}}{s_n}
\lt \dfrac{a_{n+1}}{\sum_{k=0}^{m-1} a_{n-k}}
\to \dfrac1{m}
$
so that
$r_n-1
=\dfrac{a_{n+1}}{s_n}
\to 0
$.
Your case is
$a_n = n^4$,
and it is easy to show that
$a_{n+1}/a_n \to 0$.
This holds for
$a_n = n^p$
for any fixed $p$.
A: Very short Answer
From this What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression
we have that,
$$ \sum_{k=1}^{n} k^{4} = \frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}.$$
Therefore, By polynomial limit we have, 
$$\lim_{n\to\infty}\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}= \lim_{n\to\infty}\frac{\frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}.}{\frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}.+(n+1)^4}=1$$
A: From this result: why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$
Then $\dfrac{1^4+2^4+...+n^4+(n+1)^4}{1^4+2^4+...+n^4}=1+\dfrac{(n+1)^4}{O(n^5)}=1+O(\frac 1n)\to 1$ and so is the inverse of that.
